Finding $\frac{\partial z}{\partial x}$ for $e^z=xyz$.

Question

This is simply finding $\dfrac{\partial z}{\partial x}$.

The question is $$e^z=xyz$$ and what I did was $$\ln (e^z)=\ln(xyz)$$ $$z=\ln(xyz)$$ $$\dfrac{\partial z}{\partial x}=\frac{1}{xyz}\cdot yz = \frac{1}{x}$$

But the answer is $\dfrac{\partial z}{\partial x} = \dfrac{yz}{e^z-xy}$.

How is this answer gotten?

Answer 1

We have $\exp(z)=xyz$ so by assuming $z$ as a function of-for example- $x$ so $$z_x\exp(z)=(xyz)_x=yz+xyz_x $$

Answer 2

$$\dfrac{\partial}{\partial x}\dfrac{e^z}{z}=\dfrac{\partial}{\partial x}xy$$ $$\dfrac{z\dfrac{\partial}{\partial x}e^z-e^z\dfrac{\partial}{\partial x}z}{z^2}=y$$ $$\dfrac{\partial z}{\partial x}=\dfrac{yz^2}{e^z(z-1)}=\dfrac{z}{x(z-1)}=\dfrac{yz}{xyz-xy}=\dfrac{yz}{e^z-xy}$$

Answer 3

Using your approach. $$\begin{align}z &=\ln (xyz)\\&= \ln x +\ln y+\ln z\\\dfrac{\partial z}{\partial x} &= \dfrac 1x+\require{cancel}\cancelto0{\dfrac 1y\dfrac{\partial y}{\partial x}}+\dfrac 1z\dfrac{\partial z}{\partial x}\\(1-\dfrac 1z)\dfrac{\partial z}{\partial x}&= \dfrac 1x\\ \dfrac{\partial z}{\partial x} &= \dfrac{z}{x(z-1)}\\ &=\dfrac{yz}{xyz-xy}\\ &= \dfrac{yz}{e^z-xy} \end{align}$$

Note: This assumes $y$ is independent from $x$.